## Engineering mechanics of materials |

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Page 414

8.24 An eccentrically loaded

Euler critical loading with an eccentricity of 0.1 in. What is the deflection of the

is ...

8.24 An eccentrically loaded

**column**is subjected to a loading equal to 0.6 of theEuler critical loading with an eccentricity of 0.1 in. What is the deflection of the

**column**? You are to assume that the**column**behaves elastically. 8.25 A**column**is ...

Page 416

ratios with each equation and carefully comparing the slenderness ratio of the

end conditions discussed in Section 8.4 determine

and ...

ratios with each equation and carefully comparing the slenderness ratio of the

**column**being analyzed with the prescribed slenderness ratio range. 4.**Column**end conditions discussed in Section 8.4 determine

**column**effective lengths Le ,and ...

Page 424

Example 8.10 The allowable load P on a longleaf yellow pine

determined from the following equation, which contains a factor of safety : = 1450

1 - 1.62 x 10" tLY Provided that L/d < 21, where d is the least dimension of a ...

Example 8.10 The allowable load P on a longleaf yellow pine

**column**is to bedetermined from the following equation, which contains a factor of safety : = 1450

1 - 1.62 x 10" tLY Provided that L/d < 21, where d is the least dimension of a ...

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### Contents

Introduction | 1 |

Torsionally Loaded Members in Equilibrium | 14 |

Shear and Bending Moment in Beams | 23 |

Copyright | |

19 other sections not shown

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### Common terms and phrases

absolute maximum shear allowable stress aluminum angle of twist applied Assume axes axial force axially loaded beam shown bending cantilever beam Castigliano's second theorem circular column components compressive Compute constant construct coordinate cross section cross-sectional area cylinder deflection deformation depicted in Figure Determine diameter differential elastic curve equal equation equilibrium factor of safety flexural stress free-body diagram function given by Eq Homework Problems k-ft k-in kN-m length longitudinal material maximum shear stress modulus of elasticity Mohr's circle neutral axis normal stress obtained perpendicular plane stress plot principal centroidal axis principal strains principal stresses radius Refer to Figure respect rotation section a-a segment shear center shear force shear strain shown in Figure slope solution Solve static statically indeterminate steel stress element subjected Substitution tension tion torque torsional uniform load vertical yield strength yield stress zero